15
Q:

# In an election only two candidates contested 20% of the voters did not vote and 120 votes were declared as invalid. The winner got 200 votes more than his opponent  thus he secured 41% votes of the total voters on the voter list. Percentage votes of the defeated candidate out of the total votes casted is:

 A) 47.5% B) 41% C) 38% D) 45%

Explanation:

Let there be x voters and k votes goes to loser then

0.8x - 120 = k + (k + 200)

k+200 = 0.41x

$\inline&space;\Rightarrow$  k = 1440

and  (k + 200) =1640

Therefore  $\inline&space;\frac{1440}{3200}\times&space;100$ = 45%

Q:

In a competitive examination, the average marks for the entire examination was 60 marks. If 20% of the applicants scored 85 marks and 25% scored 95 marks. What was the average marks of the remaining applicants in the examination ?

 A) 60 B) 52 C) 45 D) 35

Explanation:

Let the total applicants be 100
Then, 20% got 85 marks

i.e 20 $\fn_jvn \small \times$ 85 = 1700

and 25% got 95 marks

i.e 25 $\fn_jvn \small \times$ 95 = 2375

Now, the remaining applicants are 55 and let the average marks scored by them be x.

$\therefore$ 2375 + 1700 + 55$\fn_jvn \small \times$ x  =  60 $\fn_jvn \small \times$ 100

$\Rightarrow$ 6000 - 4075 = 55x
$\Rightarrow$ 55x=1925
$\Rightarrow$ x= $\inline \frac{1925}{55}$
$\Rightarrow$x=35.

7 27
Q:

A sales boy is allowed $\inline \fn_jvn \small 4\tfrac{1}{2}$ % discount on the total sales made by him plus a bonus of $\inline \fn_jvn \small \frac{1}{4}$% on the sales over Rs.12,000.If his total earnings were Rs.2400, then his total sales(in Rs.) are ?

 A) Rs.60,145.23 B) Rs.54,285.54 C) Rs.52,147.85 D) Rs.50,589.47

Explanation:

Let the total earnings be Rs.x

Then $\inline&space;\dpi{80}&space;\large&space;4\frac{1}{2}$% of x + $\inline&space;\dpi{80}&space;\large&space;\frac{1}{4}$% of (x-12,000) =2400

=>$\inline&space;\dpi{80}&space;\large&space;\frac{9}{2}$ $\inline \fn_jvn \small \times$ $\inline&space;\dpi{80}&space;\large&space;\frac{x}{100}$ + $\inline&space;\dpi{80}&space;\large&space;\frac{1}{4}$ $\inline \fn_jvn \small \times$ $\inline&space;\dpi{80}&space;\large&space;\frac{(x-12,000)}{100}$ =2400

=>  x=50,589.47

5 150
Q:

The population of a city is 415600.It increased by 25% in the first year and decreased by 30% in the second year.What is the population of the city at the end of second year?

As per given data, P = 415600, R1= 25% incresed, R2 = 30% decreased

Population of the city at the end of the second year= $\inline \fn_jvn p\left ( 1+\frac{R1}{100} \right )\left ( 1-\frac{R2}{100} \right )$

$\inline \fn_jvn \Rightarrow$$\inline \fn_jvn 415600\left ( 1+\frac{25}{100} \right )\left ( 1-\frac{30}{100} \right )$

$\inline \fn_jvn \Rightarrow$363650.

123
Q:

The ratio of earnings of A and B is 4:5. If the earnings of A increase by 20% and the earnings of B decrease by 20%, the new ratio of their earnings becomes 6:5. What are A's earnings?

 A) Rs.22,000 B) Rs.27,500 C) Rs. 26,400 D) Cannot be determined

Explanation:

57 2144
Q:

If the length of a rectangle is increased by 10% and the area is unchanged , then its corresponding breadth must be decreased by?

Suppose length = 100m and breadth = x m

Area = 100 m

New length = 110m and breadth = (x-y% of x)

Then, $\inline&space;110\times&space;(x-\frac{y}{100}x)=100\times&space;x$

$\inline&space;\Rightarrow$ $\inline&space;110\times&space;(1-\frac{y}{100})=100$

$\inline&space;\Rightarrow$ $\inline&space;1-\frac{y}{100}=\frac{100}{110}$

$\inline&space;\Rightarrow&space;\frac{y}{100}=1-\frac{100}{110}=\frac{1}{11}$

$\inline&space;y=\frac{100}{11}=9\frac{1}{10}$%